Bonjour,
f(x) = x² -3x-1 et g(x) = 1 + 2/(x-3)
x²-3x-1= 1 + 2/(x-3) ; x≠ 3
x²-3x-1- (1+ (2/x-3))= 0
x²-3x-1-1- (2/(x-3))= 0
x²-3x-2- (2/(x-3))= 0
[ (x-3) (x² - 3x - 2) ] -2 ]/(x-3)= 0
[ (x³ -3x²-2x-3x²+9x+6)-2 ] / (x-3)= 0
(x³-6x² + 7x+4)/(x-3)= 0
x³ -4x²-2x²+ 8x-x + 4= 0
x²(x-4) -2x(x-4) - (x-4)= 0 **** on a factorisé par (x-4)
(x-4)(x²-2x-1)= 0
on résout:
x-4= 0
x= 4
ou
x²-2x-1= 0
Δ= b²-4ac= (-2)²-4(1)(-1)= 4+4= 8 donc Δ > 0; 2 solutions
x1= (-(-2)-√8)/2(1)= (2-√8)/2= (2-√(2x4))/2= (2-2√2)/2= 1-√2
ou
x2= (-(-2)+√8)/2(1)= (2+2√2)/2= 1+√2
x= 4 ou x= 1-√2 ou x= 1+√2
S= { 4; 1-√2; 1+√2 }